Using the free energy of a monatomic ideal gas, derive

Lucia Grimes 2022-07-13 Answered
Using the free energy of a monatomic ideal gas, derive the ideal gas law
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Answers (1)

toriannucz
Answered 2022-07-14 Author has 16 answers
The entropy of the system in terms of the partition function and free energy is,
S = E T + k B log e Z
Thus, the Helmholtz free energy is,
F = E S T
Substituting the known values,
F = E ( E T + k B log e Z ) T = k B T × log e Z
The partition function for the monoatomic ideal gas is,
Z = e N ( V N ) N ( 2 π m h 2 β ) ( 2 3 N )
Substituting the known values,
F = k B T × log e ( e N ( V N ) N ( 2 π m h 2 β ) ( 3 2 N ) ) = k B T × ( log e ( e N ) + log e ( V N ) N + log e ( 2 π m h 2 β ) 3 2 N ) = k B T × ( N + N log e ( V N ) + ( 3 2 N ) log e ( 2 π m h 2 β ) 1 )
As the relation between pressure and the Helmholtz free energy is,
P = ( d F d V ) T , N = k B T N V
As the gas constant is defined as,
R = N A k B k B = R N A
where N A is an Avogadro's number and k B is a Boltzmann constant
Thus, the pressure equation become,
P = R T N N A V P V = N N A R T P V = n a R T
where n A is the number of moles of gas.
This equation represents the ideal gas equation.

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